Integrand size = 22, antiderivative size = 222 \[ \int \frac {\sqrt {1+2 x}}{2+3 x+5 x^2} \, dx=-\sqrt {\frac {2}{5 \left (-2+\sqrt {35}\right )}} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\sqrt {\frac {2}{5 \left (-2+\sqrt {35}\right )}} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\frac {\log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{\sqrt {10 \left (2+\sqrt {35}\right )}}-\frac {\log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{\sqrt {10 \left (2+\sqrt {35}\right )}} \]
-1/5*arctan((-10*(1+2*x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+10*35^(1/2))^( 1/2))*10^(1/2)/(-2+35^(1/2))^(1/2)+1/5*arctan((10*(1+2*x)^(1/2)+(20+10*35^ (1/2))^(1/2))/(-20+10*35^(1/2))^(1/2))*10^(1/2)/(-2+35^(1/2))^(1/2)+ln(5+1 0*x+35^(1/2)-(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))/(20+10*35^(1/2))^(1/2)- ln(5+10*x+35^(1/2)+(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))/(20+10*35^(1/2))^ (1/2)
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {1+2 x}}{2+3 x+5 x^2} \, dx=\frac {2 \left (\sqrt {2-i \sqrt {31}} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )+\sqrt {2+i \sqrt {31}} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )\right )}{\sqrt {155}} \]
(2*(Sqrt[2 - I*Sqrt[31]]*ArcTan[Sqrt[(-2 - I*Sqrt[31])/7]*Sqrt[1 + 2*x]] + Sqrt[2 + I*Sqrt[31]]*ArcTan[Sqrt[(I/7)*(2*I + Sqrt[31])]*Sqrt[1 + 2*x]])) /Sqrt[155]
Time = 0.53 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1148, 1447, 27, 1475, 1083, 217, 1478, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt {2 x+1}}{5 x^2+3 x+2} \, dx\) |
| \(\Big \downarrow \) 1148 |
| \(\displaystyle 4 \int \frac {2 x+1}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\) |
| \(\Big \downarrow \) 1447 |
| \(\displaystyle 4 \left (\frac {1}{2} \int \frac {5 (2 x+1)+\sqrt {35}}{5 \left (5 (2 x+1)^2-4 (2 x+1)+7\right )}d\sqrt {2 x+1}-\frac {1}{2} \int \frac {\sqrt {35}-5 (2 x+1)}{5 \left (5 (2 x+1)^2-4 (2 x+1)+7\right )}d\sqrt {2 x+1}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {1}{10} \int \frac {5 (2 x+1)+\sqrt {35}}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}-\frac {1}{10} \int \frac {\sqrt {35}-5 (2 x+1)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\right )\) |
| \(\Big \downarrow \) 1475 |
| \(\displaystyle 4 \left (\frac {1}{10} \left (\frac {1}{2} \int \frac {1}{2 x-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {\frac {7}{5}}+1}d\sqrt {2 x+1}+\frac {1}{2} \int \frac {1}{2 x+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {\frac {7}{5}}+1}d\sqrt {2 x+1}\right )-\frac {1}{10} \int \frac {\sqrt {35}-5 (2 x+1)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle 4 \left (\frac {1}{10} \left (-\int \frac {1}{-2 x+\frac {2}{5} \left (2-\sqrt {35}\right )-1}d\left (2 \sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )-\int \frac {1}{-2 x+\frac {2}{5} \left (2-\sqrt {35}\right )-1}d\left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )-\frac {1}{10} \int \frac {\sqrt {35}-5 (2 x+1)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle 4 \left (\frac {1}{10} \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )+\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )\right )-\frac {1}{10} \int \frac {\sqrt {35}-5 (2 x+1)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}\right )\) |
| \(\Big \downarrow \) 1478 |
| \(\displaystyle 4 \left (\frac {1}{10} \left (\frac {1}{2} \sqrt {\frac {5}{2 \left (2+\sqrt {35}\right )}} \int -\frac {5 \left (\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-2 \sqrt {2 x+1}\right )}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{2} \sqrt {\frac {5}{2 \left (2+\sqrt {35}\right )}} \int -\frac {5 \left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )+\frac {1}{10} \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )+\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {1}{10} \left (-\frac {5}{2} \sqrt {\frac {5}{2 \left (2+\sqrt {35}\right )}} \int \frac {\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-2 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {5}{2} \sqrt {\frac {5}{2 \left (2+\sqrt {35}\right )}} \int \frac {2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )+\frac {1}{10} \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )+\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )\right )\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle 4 \left (\frac {1}{10} \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )+\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \arctan \left (\sqrt {\frac {5}{2 \left (\sqrt {35}-2\right )}} \left (2 \sqrt {2 x+1}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )\right )\right )+\frac {1}{10} \left (\frac {1}{2} \sqrt {\frac {5}{2 \left (2+\sqrt {35}\right )}} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )-\frac {1}{2} \sqrt {\frac {5}{2 \left (2+\sqrt {35}\right )}} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )\right )\right )\) |
4*((Sqrt[5/(2*(-2 + Sqrt[35]))]*ArcTan[Sqrt[5/(2*(-2 + Sqrt[35]))]*(-Sqrt[ (2*(2 + Sqrt[35]))/5] + 2*Sqrt[1 + 2*x])] + Sqrt[5/(2*(-2 + Sqrt[35]))]*Ar cTan[Sqrt[5/(2*(-2 + Sqrt[35]))]*(Sqrt[(2*(2 + Sqrt[35]))/5] + 2*Sqrt[1 + 2*x])])/10 + ((Sqrt[5/(2*(2 + Sqrt[35]))]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt [35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/2 - (Sqrt[5/(2*(2 + Sqrt[35]))]*Log[S qrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/2)/10)
3.24.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[Sqrt[(d_.) + (e_.)*(x_)]/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[2*e Subst[Int[x^2/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c *x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a/c, 2]}, Simp[1/2 Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b ^2 - 4*a*c, 0] && PosQ[a*c]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^ 2, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[-2*(d/e) - b/c, 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ [c*d^2 - a*e^2, 0] && !GtQ[b^2 - 4*a*c, 0]
Time = 0.52 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.09
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (\frac {\left (\sqrt {5}-\frac {5 \sqrt {7}}{2}\right ) \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{310}-\frac {\left (\sqrt {5}-\frac {5 \sqrt {7}}{2}\right ) \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{310}+\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(241\) |
| derivativedivides | \(-\frac {\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \left (2 \sqrt {5}-5 \sqrt {7}\right ) \left (\frac {\ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{10}+\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{31}+\frac {\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \left (2 \sqrt {5}-5 \sqrt {7}\right ) \left (\frac {\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{10}-\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{31}\) | \(270\) |
| default | \(-\frac {\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \left (2 \sqrt {5}-5 \sqrt {7}\right ) \left (\frac {\ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{10}+\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{31}+\frac {\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \left (2 \sqrt {5}-5 \sqrt {7}\right ) \left (\frac {\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{10}-\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{31}\) | \(270\) |
| trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}+620\right ) \ln \left (\frac {57660 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}+620\right ) \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{4} x -155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}+620\right ) x -2976 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}+620\right )-14415 \sqrt {1+2 x}\, \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}-55 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}+620\right ) x -88 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}+620\right )-4495 \sqrt {1+2 x}}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2} x +5 x +4}\right )}{155}-\operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right ) \ln \left (\frac {288300 x \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{5}+15655 x \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{3}+14880 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{3}+465 \sqrt {1+2 x}\, \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2}-63 x \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )-56 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )-133 \sqrt {1+2 x}}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}+124 \textit {\_Z}^{2}+7\right )^{2} x -x -4}\right )\) | \(421\) |
-2*(1/310*(5^(1/2)-5/2*7^(1/2))*(10*5^(1/2)*7^(1/2)-20)^(1/2)*(2*5^(1/2)*7 ^(1/2)+4)^(1/2)*ln(5^(1/2)*7^(1/2)-(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+ 2*x)^(1/2)+5+10*x)-1/310*(5^(1/2)-5/2*7^(1/2))*(10*5^(1/2)*7^(1/2)-20)^(1/ 2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)*ln(5^(1/2)*7^(1/2)+(2*5^(1/2)*7^(1/2)+4)^(1 /2)*5^(1/2)*(1+2*x)^(1/2)+5+10*x)+arctan((5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1 /2)-10*(1+2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))-arctan((5^(1/2)*(2*5^ (1/2)*7^(1/2)+4)^(1/2)+10*(1+2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2)))/( 10*5^(1/2)*7^(1/2)-20)^(1/2)
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {1+2 x}}{2+3 x+5 x^2} \, dx=\frac {1}{310} \, \sqrt {155} \sqrt {4 i \, \sqrt {31} - 8} \log \left (i \, \sqrt {155} \sqrt {31} \sqrt {4 i \, \sqrt {31} - 8} + 310 \, \sqrt {2 \, x + 1}\right ) - \frac {1}{310} \, \sqrt {155} \sqrt {4 i \, \sqrt {31} - 8} \log \left (-i \, \sqrt {155} \sqrt {31} \sqrt {4 i \, \sqrt {31} - 8} + 310 \, \sqrt {2 \, x + 1}\right ) - \frac {1}{310} \, \sqrt {155} \sqrt {-4 i \, \sqrt {31} - 8} \log \left (i \, \sqrt {155} \sqrt {31} \sqrt {-4 i \, \sqrt {31} - 8} + 310 \, \sqrt {2 \, x + 1}\right ) + \frac {1}{310} \, \sqrt {155} \sqrt {-4 i \, \sqrt {31} - 8} \log \left (-i \, \sqrt {155} \sqrt {31} \sqrt {-4 i \, \sqrt {31} - 8} + 310 \, \sqrt {2 \, x + 1}\right ) \]
1/310*sqrt(155)*sqrt(4*I*sqrt(31) - 8)*log(I*sqrt(155)*sqrt(31)*sqrt(4*I*s qrt(31) - 8) + 310*sqrt(2*x + 1)) - 1/310*sqrt(155)*sqrt(4*I*sqrt(31) - 8) *log(-I*sqrt(155)*sqrt(31)*sqrt(4*I*sqrt(31) - 8) + 310*sqrt(2*x + 1)) - 1 /310*sqrt(155)*sqrt(-4*I*sqrt(31) - 8)*log(I*sqrt(155)*sqrt(31)*sqrt(-4*I* sqrt(31) - 8) + 310*sqrt(2*x + 1)) + 1/310*sqrt(155)*sqrt(-4*I*sqrt(31) - 8)*log(-I*sqrt(155)*sqrt(31)*sqrt(-4*I*sqrt(31) - 8) + 310*sqrt(2*x + 1))
\[ \int \frac {\sqrt {1+2 x}}{2+3 x+5 x^2} \, dx=\int \frac {\sqrt {2 x + 1}}{5 x^{2} + 3 x + 2}\, dx \]
\[ \int \frac {\sqrt {1+2 x}}{2+3 x+5 x^2} \, dx=\int { \frac {\sqrt {2 \, x + 1}}{5 \, x^{2} + 3 \, x + 2} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (162) = 324\).
Time = 0.70 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.08 \[ \int \frac {\sqrt {1+2 x}}{2+3 x+5 x^2} \, dx=\frac {1}{37215500} \, \sqrt {31} {\left (210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} - \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )}\right )} \arctan \left (\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{37215500} \, \sqrt {31} {\left (210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} - \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )}\right )} \arctan \left (-\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} - \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{74431000} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )} - 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}}\right )} \log \left (2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + 2 \, x + \sqrt {\frac {7}{5}} + 1\right ) - \frac {1}{74431000} \, \sqrt {31} {\left (\sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )} - 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}}\right )} \log \left (-2 \, \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + 2 \, x + \sqrt {\frac {7}{5}} + 1\right ) \]
1/37215500*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140* sqrt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 2*( 7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35))*arctan(5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35* sqrt(35) + 1/2) + sqrt(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2)) + 1/37215500* sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 2*(7/5)^(3/4)* (140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2 *sqrt(35) - 35))*arctan(-5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35*sqrt(35) + 1/2) - sqrt(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2)) + 1/74431000*sqrt(31)*( sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 210*sqrt(31)*(7/5)^(3/4 )*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 420*(7/5)^(3/4)*(2*sqrt(35 ) + 35)*sqrt(-140*sqrt(35) + 2450) + 2*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^ (3/2))*log(2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(35) + 1/2) + 2*x + s qrt(7/5) + 1) - 1/74431000*sqrt(31)*(sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 210*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(3 5) - 35) - 420*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) + 2*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2))*log(-2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(35) + 1/2) + 2*x + sqrt(7/5) + 1)
Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {1+2 x}}{2+3 x+5 x^2} \, dx=-\frac {2\,\sqrt {155}\,\mathrm {atanh}\left (\sqrt {155}\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}\,\left (\frac {2\,\left (\frac {2}{155}+\frac {\sqrt {31}\,1{}\mathrm {i}}{155}\right )\,\sqrt {2\,x+1}}{7}+\frac {27\,\sqrt {2\,x+1}}{1085}\right )\right )\,\sqrt {-2-\sqrt {31}\,1{}\mathrm {i}}}{155}-\frac {2\,\sqrt {155}\,\mathrm {atanh}\left (-\sqrt {155}\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}\,\left (\frac {2\,\left (-\frac {2}{155}+\frac {\sqrt {31}\,1{}\mathrm {i}}{155}\right )\,\sqrt {2\,x+1}}{7}-\frac {27\,\sqrt {2\,x+1}}{1085}\right )\right )\,\sqrt {-2+\sqrt {31}\,1{}\mathrm {i}}}{155} \]
- (2*155^(1/2)*atanh(155^(1/2)*(- 31^(1/2)*1i - 2)^(1/2)*((2*((31^(1/2)*1i )/155 + 2/155)*(2*x + 1)^(1/2))/7 + (27*(2*x + 1)^(1/2))/1085))*(- 31^(1/2 )*1i - 2)^(1/2))/155 - (2*155^(1/2)*atanh(-155^(1/2)*(31^(1/2)*1i - 2)^(1/ 2)*((2*((31^(1/2)*1i)/155 - 2/155)*(2*x + 1)^(1/2))/7 - (27*(2*x + 1)^(1/2 ))/1085))*(31^(1/2)*1i - 2)^(1/2))/155